1. Field of the Invention
This invention relates to power supplies and more specifically, to power supplies which generate a square wave from a DC input (inverters).
2. Description of the Prior Art
Inverters receive a DC voltage and generate therefrom a square wave output voltage. FIG. 1 illustrates an inverter 10 which performs this function. Inverter 10 includes a pair of terminals 12a and 12b, across which a DC voltage (typically, about 160 volts) is applied. Inverter 10 includes a power transistor Q1 and a power transistor Q2. In operation, power transistor Q1 is on while power transistor Q2 is off, and vice versa. This causes a square wave voltage to be provided across the primary winding T1P of a first transformer T1. This in turn causes a square wave output voltage to be generated across the secondary winding T1S of transformer T1. The period for the square wave provided by inverter 10 is determined by the characteristics of a second transformer T2. Second transformer T2 has three windings: a set of windings T2A, T2B and T2C.
In operation, when transistor Q1 is on, current flows from the emitter of transistor Q1 through winding T2A and through primary winding T1P of transformer T1. When that happens, a current also flows through winding T2B of transformer T2. Typically, winding T2A has one turn while winding T2B has eight turns. Therefore, if a current I.sub.T2A flows through winding T2A, then a current I.sub.T2B =I.sub.T2A /8 will flow through winding T2B. Current I.sub.T2B then flows through a diode D1 and into the base of transistor Q1. The voltage across winding T2B equals the voltage across diode D1 plus the voltage across the base emitter junction of transistor Q1. Typically, this voltage equals approximately 1.4 volts. A capacitor C1 is connected across diode D1. Therefore, while transistor Q1 is on, a voltage of one diode drop, or approximately 0.7 volts, is applied across the terminals C1A and C1B of capacitor C1, thus charging capacitor C1 to 0.7 volts.
As is well understood in the art, when a voltage is applied across a saturable core, such as the core associated with winding T2B, eventually the saturable core saturates. When that happens, the voltage across winding T2B drops to zero. Because of this, a reverse current will flow through diode D1 and the base of transistor Q1 (because of charge stored in the PN junction of diode D1), causing transistor Q1 to turn off. Further, because capacitor C1 has been charged to a voltage of 0.7 volts, when the voltage on terminal C1A of capacitor C1 drops to a potential equal to the potential of the emitter of transistor Q1, the voltage on terminal C1B of capacitor C1 drops to a voltage of -0.7 volts relative to the emitter voltage of transistor Q1. This turns off transistor Q1 even more rapidly. Because the reverse current out of the base of transistor Q1 falls to zero, and current can no longer flow through winding T2B of transformer T2 because transistor Q1 is shut off, the flux stored in transformer T2 causes a current to start flowing through winding T2C. When this happens, current flows in the direction of arrow A through a diode D2 and through the base of transistor Q2, causing transistor Q2 to turn on. The voltage across winding T2C equals the voltage drop across diode D2 plus the base emitter junction voltage drop across the base emitter junction of transistor Q2, or approximately 1.4 volts. Because the voltage across winding T2C equals 1.4 volts, the voltage across winding T2B equals -1.4 volts (the core of winding T2B is no longer saturated).
When transistor Q2 turns on, current flows through winding T2A and through the collector of transistor Q2. When transistor Q2 is on, the voltage across primary winding T1P of transformer T1 equals the negative of the voltage that appears across primary winding T1P when transistor Q1 is on. Transistor Q2 will remain on until the core of transformer T2 goes back into saturation. Specifically, since the voltage of 1.4 volts is constantly applied across winding T2C, eventually transformer T2 goes back into saturation in a polarity opposite that which occurs when transistor Q1 turns off. When that happens, zero volts appear across winding T2C and transistor Q2 turns off. A voltage then appears across winding T2B and a current sufficient to turn on transistor Q1 starts to flow.
It is seen that half the period of oscillation for inverter 10 is determined by the length of time it takes transformer T2 to go from positive to negative saturation, and the other half period of oscillation is determined by the lenght of time it takes transformer T2 to go from negative to positive saturation. This in turn is fixed, at least in part, by the physical dimensions of transformer T2 and the magnitude of the voltage appearing across windings T2B and T2C.
The voltages appearing across winding T2B and winding T2C are determined by the voltage drop across diode D1 and the base emitter junction of transistor Q1 and the voltage drop across diode D2 and the voltage drop across the base emitter junction of transistor Q2, respectively. Because diode D1 is not a perfect diode, and transistor Q1 is not a perfect transistor, some resistance is associated with these elements. Therefore, an increase in the current flowing through winding T2B causes an increase in the voltage drop across diode D1 and across the base emitter junction of transistor Q1. Similarly, an increase in the current across transformer winding T2C, causes the voltage drop across diode D2 and the voltage drop across the base emitter junction of transistor Q2 to increase. Thus, a large current through winding T2B and winding T2C (causing a larger voltage across these windings) will cause transformer T2 to go into saturation more rapidly than a small current through windings T2B and T2C. The current through winding T2B and winding T2C depends on the current flowing through winding T2A which, in turn, is determined by the current through primary winding T1P of transformer T1 which, in turn, depends upon the load current out of secondary winding T1S of transformer T1. It is thus seen that the greater the load current for inverter 10, the greater the frequency of oscillation of inverter 10. For various applications, this is undesirable. In fact, for some applications, it is desirable to have a slightly lower frequency and a slightly longer period when the load current is increased.